Is ${493047}$ divisible by $9$ ?
Explanation: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {493047}= &&{4}\cdot100000+ \\&&{9}\cdot10000+ \\&&{3}\cdot1000+ \\&&{0}\cdot100+ \\&&{4}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {493047}= &&{4}(99999+1)+ \\&&{9}(9999+1)+ \\&&{3}(999+1)+ \\&&{0}(99+1)+ \\&&{4}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {493047}= &&\gray{4\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{4\cdot9}+ \\&& {4}+{9}+{3}+{0}+{4}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${493047}$ is divisible by $9$ if ${ 4}+{9}+{3}+{0}+{4}+{7}$ is divisible by $9$ Add the digits of ${493047}$ $ {4}+{9}+{3}+{0}+{4}+{7} = {27} $ If ${27}$ is divisible by $9$ , then ${493047}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${493047}$ must also be divisible by $9$.